Judge the parity of the following functions: ① f (x) = √ X-1 + √ 1-x; ② f (x) = | x | + √ X & # 178; ③ f (x) = √ 1-x & # 178; / | x + 2 | - 3 The process should be detailed

Judge the parity of the following functions: ① f (x) = √ X-1 + √ 1-x; ② f (x) = | x | + √ X & # 178; ③ f (x) = √ 1-x & # 178; / | x + 2 | - 3 The process should be detailed

(1) Because the function domain is {1}, when x belongs to the domain, - X does not belong to the domain
(2) Even function. The domain of definition is R. f (x) = | x | + | x | = 2 | x |, f (- x) = 2 | - x | = 2 | - x | = f (x), so the function is even
(3) Because f (1 / 2) = √ (3 / 4) / (- 1 / 2), f (- 1 / 2) = √ (3 / 4) / (- 3 / 2), they are neither equal nor opposite
Judge the parity of the following functions: 1, f (x) = 2x & # 178; + 4, X ∈ (- 2,2) 2, f (x) = | 2x-1 | - | 2x + 1|
f(x)=2x²﹢4
f(-x)=2(-x)²﹢4=2x²﹢4=f(x)
So f (x) is even function
f﹙x﹚=|2x－1|－|2x+1|
f﹙-x﹚=|-2x－1|－|-2x+1|=|2x+1|－|2x-1|=-f(x)
So f (x) is an odd function
Judge the parity of the following functions (1) f (x) = | 2x + 1 | + | 2x-1 |, (2) f (x) = / X √ x2-4
Given the set a = {x | AX3 x2-x = 0}, if the set a is a single element set, then the value range of real number a is?
In set a is AX3 + x2-x = 0!
When x = 0, the value range of a is from negative infinity to positive infinity
If x is not equal to 0, then AX2 + X-1 = 0
Because x is unique, the above equation is unique, then a = - 1 / 4, then x = 2
Finally, summarize the scope yourself
It's very simple. Set a contains element zero, so when x is not equal to 0, AX2 + X-1 = 0 has no real number root
If the function y equals 2-m + (the square of M-4) is an inverse proportional function of X, then M equals
m=-2
∵ is the inverse scale function of Y with respect to X
∴m²-4=0
∴2-m≠0
∴m=-2
The answer is - 2 percent
Ten
If the inverse scale function is used, the degree of X is - 1, so the coefficient of 2-m 2; = - 1 m 2; = 3 M = ± √ 3 in the second and fourth quadrants is less than 0 m + 1
Let the elements in the set a be real numbers. When a belongs to a, 1 / 1-A belongs to A. (1) prove that if a belongs to a, then 1-1 / a belongs to A. (2) if 2 belongs to a, find the set
）Because when a belongs to a, 1 / 1-A belongs to a, obviously a cannot be equal to 1
Then 1 / 1-A also belongs to a, so 1 / 1 - (1 / 1-A) = 1-1 / A
How did 1 / 1 - (1 / 1-A) = 1-1 / a change?
When a belongs to a, 1 / 1-A belongs to a
Because 1 / 1-A also belongs to a, we substitute 1 / 1-A into a
The monotone increasing interval of the function y = 2tan (3x + π 4) - 5 is
=Monotone increasing interval of 2tan (3x + π / 4)
-π/2+kπ
For the set a whose elements are real numbers, if a belongs to a, then (1 + a) / (1-A) belongs to a
For the set a whose elements are real numbers, if a belongs to a, then (1 + a) / (1-A) belongs to a
1. Given that 2 belongs to a, find the set a
2. Try to find a number B, make B belong to a, and find out the set a
3. According to the known conditions and the results of questions 1 and 2, what conclusions can we draw? Write two without proof
Better be clear and clear
1. 2 ∈ a, then
(1＋2)/(1－2)＝－3∈A
(1－3)/(1＋3)＝－1/2∈A
(1－1/2)/(1＋1/2)＝1/3∈A
Because (1 + 1 / 3) / (1-1 / 3) = 2, the elements in a are 2, - 3, - 1 / 2, 1 / 3, that is, a = {2, - 3, - 1 / 2, 1 / 3}
2. If a ∈ a, (1 + a) / (1-A) ∈ a, then
[1＋(1＋a)/(1－a)]/[1－(1＋a)/(1－a)]＝－1/a∈A
[1＋(－1/a)]/[1－(－1/a)]＝(a－1)/(a＋1)∈A
And [1 + (A-1) / (a + 1)] / [1 - (A-1) / (a + 1)] = a, so
A＝{a,－1/a,(1＋a)/(1－a),(a－1)/(a＋1)}
3. Conclusion 1:1 and 0 are not in set a
Conclusion 2: the elements in a appear in pairs, and the product of the two elements is - 1
（1+2）/（1-2）=-3
（1-3）/（1+3）=-1/2
（1-1/2）/（1+1/2）=1/3
（1+1/3）/（1-1/3）=2
……
A =｛-3，-1/2,1/3,2,｝
None in a
The monotone increasing interval of function y = 2tan (3x + π 4) - 5 is - 5!
-5 has no effect on finding monotone increasing interval
Monotone increasing interval of y = 2tan (3x + π / 4)
-π/2+kπ
It is known that the elements of set a are real numbers and satisfy the following conditions: 1.1 does not belong to A. 2. If a belongs to a, then 1 / 1-A belongs to a
Ask for:
1. If a belongs to a, try to find other elements of set a
2. Can set a be a single element set? If so, find out the elements of the set. If not, explain the reason
1.1/1-a belongs to a, so there is no need to analyze it. Then there is also a relative 1 / 1-A, that is, 1 / (1-1 / 1-A), that is, A-1 / A, which is substituted to get 1 / [1 - (A-1) / a], and the answer is a, so there is no need to do it any more. So there are three elements in a set. One is omitted above